Let z be a complex number such that |z + 2| = | vedclass

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(A) Given $|z + 2| = |z - 2|$, this represents the perpendicular bisector of the segment joining $-2$ and $2$, which is the imaginary axis. Thus, $z =

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$9$

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(A) Given $|z + 2| = |z - 2|$, this represents the perpendicular bisector of the segment joining $-2$ and $2$, which is the imaginary axis. Thus, $z = iy$ for some $y \in \mathbb{R}$.Substituting $z = iy$ into the argument expression:$\frac{z + 3}{z - i} = \frac{3 + iy}{iy - i} = \frac{3 + iy}{i(y - 1)}$.To simplify, multiply the numerator and denominator by $-i$:$\frac{(3 + iy)(-i)}{i(y - 1)(-i)} = \frac{-3i - i^2y}{y - 1} = \frac{y - 3i}{y - 1} = \frac{y}{y - 1} - i\frac{3}{y - 1}$.Given $\arg\left(\frac{z + 3}{z - i}ight) = \frac{\pi}{4}$, we have $	an\left(\frac{\pi}{4}ight) = \frac{	ext{Im}}{	ext{Re}} = \frac{-3/(y - 1)}{y/(y - 1)} = \frac{-3}{y} = 1$.Solving for $y$, we get $y = -3$.Thus, $z = -3i$, and $|z|^2 = |-3i|^2 = 9$.

Let $z$ be a complex number such that $|z + 2| = |z - 2|$ and $\arg\left(\frac{z + 3}{z - i}\right) = \frac{\pi}{4}$. Then $|z|^2$ is equal to:

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